Questions: Question 24 (1 point) State the null and alternative hypotheses. A study was conducted to determine if there is a relationship between fan preference of instant replay use and the sport in which it is applied. The category counts of 102 fans are provided in the two-way table below. Use a ChiSquare independence test to determine if fan preference of instant replay use and the sport in which it is applied are independent at the 5% level of significance. State the Null and Alternative Hypotheses. H0: Instant replay preference is independent of sport. HA: Instant replay preference is dependent on sport.

Question 24 (1 point)
State the null and alternative hypotheses.
A study was conducted to determine if there is a relationship between fan preference of instant replay use and the sport in which it is applied. The category counts of 102 fans are provided in the two-way table below. Use a ChiSquare independence test to determine if fan preference of instant replay use and the sport in which it is applied are independent at the 5% level of significance.

State the Null and Alternative Hypotheses.
H0: Instant replay preference is independent of sport.
HA: Instant replay preference is dependent on sport.

Transcript text: Question 24 (1 point) State the null and alternative hypotheses. A study was conducted to determine if there is a relationship between fan preference of instant replay use and the sport in which it is applied. The category counts of 102 fans are provided in the two-way table below. Use a ChiSquare independence test to determine if fan preference of instant replay use and the sport in which it is applied are independent at the $5 \%$ level of significance. State the Null and Alternative Hypotheses. $H_{0}$ : Instant replay preference is independent of sport. $H_{A}$: Instant replay preference is dependent on sport.

Solution

Solution Steps

Step 1: State the Hypotheses

We define the null and alternative hypotheses as follows:

Null Hypothesis ($H_0$): Instant replay preference is independent of sport.
Alternative Hypothesis ($H_A$): Instant replay preference is dependent on sport.

Step 2: Calculate Expected Frequencies

Using the formula for expected frequency $E$ for each cell in the contingency table:

\[ E = \frac{R_i \times C_j}{N} \]

where $R_i$ is the total for row $i$, $C_j$ is the total for column $j$, and $N$ is the total number of observations.

The expected frequencies are calculated as follows:

For cell (1, 1): \[ E = \frac{24 \times 57}{102} = 13.4118 \]
For cell (1, 2): \[ E = \frac{24 \times 45}{102} = 10.5882 \]
For cell (2, 1): \[ E = \frac{24 \times 57}{102} = 13.4118 \]
For cell (2, 2): \[ E = \frac{24 \times 45}{102} = 10.5882 \]
For cell (3, 1): \[ E = \frac{41 \times 57}{102} = 22.9118 \]
For cell (3, 2): \[ E = \frac{41 \times 45}{102} = 18.0882 \]
For cell (4, 1): \[ E = \frac{13 \times 57}{102} = 7.2647 \]
For cell (4, 2): \[ E = \frac{13 \times 45}{102} = 5.7353 \]

The expected frequencies are: \[ \begin{bmatrix} 13.4118 & 10.5882 \\ 13.4118 & 10.5882 \\ 22.9118 & 18.0882 \\ 7.2647 & 5.7353 \end{bmatrix} \]

Step 3: Calculate Chi-Square Test Statistic

The Chi-Square test statistic ($\chi^2$) is calculated using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where $O$ is the observed frequency and $E$ is the expected frequency. The calculations for each cell are as follows:

For cell (1, 1): \[ O = 19, E = 13.4118 \Rightarrow \frac{(19 - 13.4118)^2}{13.4118} = 2.3284 \]
For cell (1, 2): \[ O = 5, E = 10.5882 \Rightarrow \frac{(5 - 10.5882)^2}{10.5882} = 2.9493 \]
For cell (2, 1): \[ O = 18, E = 13.4118 \Rightarrow \frac{(18 - 13.4118)^2}{13.4118} = 1.5697 \]
For cell (2, 2): \[ O = 6, E = 10.5882 \Rightarrow \frac{(6 - 10.5882)^2}{10.5882} = 1.9882 \]
For cell (3, 1): \[ O = 15, E = 22.9118 \Rightarrow \frac{(15 - 22.9118)^2}{22.9118} = 2.7320 \]
For cell (3, 2): \[ O = 26, E = 18.0882 \Rightarrow \frac{(26 - 18.0882)^2}{18.0882} = 3.4606 \]
For cell (4, 1): \[ O = 5, E = 7.2647 \Rightarrow \frac{(5 - 7.2647)^2}{7.2647} = 0.7060 \]
For cell (4, 2): \[ O = 8, E = 5.7353 \Rightarrow \frac{(8 - 5.7353)^2}{5.7353} = 0.8943 \]

Summing these values gives: \[ \chi^2 = 2.3284 + 2.9493 + 1.5697 + 1.9882 + 2.7320 + 3.4606 + 0.7060 + 0.8943 = 16.6286 \]

Step 4: Determine Critical Value and P-Value

The critical value for $\chi^2$ at $\alpha = 0.05$ with 3 degrees of freedom is: \[ \chi^2_{\alpha, df} = 7.8147 \]

The p-value associated with the calculated $\chi^2$ statistic is: \[ P = P(\chi^2 > 16.6286) = 0.0008 \]

Step 5: Conclusion

Since the calculated $\chi^2$ statistic $16.6286$ is greater than the critical value $7.8147$ and the p-value $0.0008$ is less than $\alpha = 0.05$, we reject the null hypothesis.